Difference between revisions of "1966 AHSME Problems/Problem 33"
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== Solution == | == Solution == | ||
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Let <math>m=\frac{x-a}{b}</math> and <math>n=\frac{x-b}{a}</math> then we have | Let <math>m=\frac{x-a}{b}</math> and <math>n=\frac{x-b}{a}</math> then we have | ||
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Solving the quadratic gets another <math>2</math> solutions for <math>x</math>. | Solving the quadratic gets another <math>2</math> solutions for <math>x</math>. | ||
| − | Thus there are <math>\boxed{ | + | Thus there are <math>\boxed{\text{(D) three}}</math> solutions in total. |
~ Nafer | ~ Nafer | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1966|num-b=32|num-a=34}} | + | {{AHSME 40p box|year=1966|num-b=32|num-a=34}} |
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:26, 29 July 2024
Problem
If 
and 
, the number of distinct values of 
satisfying the equation
![\[\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b},\]](http://latex.artofproblemsolving.com/b/4/d/b4dfad7746d087a93d579bcbceadafc0c3ca8d55.png)
is:
Solution
Let 
and 
then we have
![\[m+n=\frac{1}{m}+\frac{1}{n}\]](http://latex.artofproblemsolving.com/2/d/9/2d9a16cf378036b0d90044e1501ce0fab7cac286.png)
![\[m+n=\frac{m+n}{mn}\]](http://latex.artofproblemsolving.com/c/5/2/c52338c48d698a51f3795fa1dbff72e32ee8b735.png)
Notice that the equation is possible iff 
or 
.
If then
![\[\frac{x-a}{b}+\frac{x-b}{a}=0\]](http://latex.artofproblemsolving.com/f/8/f/f8f7e51ee192b3d0d2cfa11a5fd0f2d31b14072d.png)
![\[\frac{x-a}{b}=\frac{b-x}{a}\]](http://latex.artofproblemsolving.com/d/c/7/dc7aad84f4f267e2d3799d1b76a77d61916acfc3.png)
![\[x=\frac{a^2+b^2}{a+b}\]](http://latex.artofproblemsolving.com/b/5/9/b5955801ffeab483090ee0582c532594950279d9.png)
Which yields 
solution for .
If 
then
![\[(\frac{x-a}{b})(\frac{x-b}{a})=1\]](http://latex.artofproblemsolving.com/7/4/e/74ea230cad26c7f2ea991f06989c202347f330b0.png)
![\[x^2-(a+b)x=0\]](http://latex.artofproblemsolving.com/8/2/6/8267efa0ad146faab8a61381c566c442f9a532a8.png)
Solving the quadratic gets another 
solutions for .
Thus there are 
solutions in total.
~ Nafer
See also
| 1966 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 32 |
Followed by Problem 34 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
